Physics Alternating Current

Here $?=45°$ and current leads voltage, so $X_C>X_L$

$tan?=\frac{X}{R}=tan45°=\text{?}1\Rightarrow X=X_C-X_L=R$

$\Rightarrow X_C=R+L\omega \Rightarrow X_C=1+30*10^{-3}*300=10$

$\Rightarrow \frac{1}{C\omega }=10\Rightarrow C=\frac{1}{10*\omega }=\frac{1}{10*300}=\frac{1}{3}*10^{-3}F$  

  $\omega =100\pi ,R=100\Omega ,X_L=L\omega =0.5*10^{-3}*100\pi =5\pi *10^{-2}\Omega ,$ and

  $X_C=\frac{1}{C\omega }=\frac{1}{0.1*10^{-12}*100\pi }=\frac{10^{11}}{\pi }\Omega$            

Since X_C is approximately infinite, so the phase angle between current and supplied voltage and the nature of the circuit is $\approx 90°,$ predominantly capacitive circuit.

$X_c=\frac{1}{\omega c},X_L=\omega L$ so at very high frequencies capacitor behaves as conduct and inductor behaves as open circuit. The effective impedance will be $2\Omega .$

$Z=\sqrt[]{R^2+{\left(X_C-X_L\right)}^2}$

$Z=\sqrt[]{R^2+{\left(R-R\right)}^2}\left[?X_L=X_C=R\to Given\right]$

Z = R

$cos{?}_1=\frac{R}{\sqrt[]{R^2+{\left(3R\right)}^2}}=\frac{1}{\sqrt[]{10}}$

$cos{?}_2=\frac{R}{\sqrt[]{R^2+{\left(3R-R\right)}^2}}=\frac{1}{\sqrt[]{2}}$

$\Rightarrow \frac{cos{?}_2}{cos{?}_1}=\frac{\sqrt[]{10}}{\sqrt[]{2}}=\sqrt[]{5}\Rightarrow x=1$

$f_{rms}^2={\left(\frac{\sqrt[]{42}}{\sqrt[]{2}}\right)}^2+10^2=121$

$\Rightarrow f_{rms}=11A$

$\Delta V_{D1}=\Delta V_{D2}=l*R=0$

Since resistance is zero in forward bias.

=> V₀ = 5V

Let, current through inductor,

$i_L=l_{L}sin\left(\omega t+{?}_1\right)$                

$l_L=\frac{200\sqrt[]{2}}{\sqrt[]{50^2+{\left(100*0.50\right)}^2}}=4A$              

${?}_1=ta{n}^{-1}\left(\frac{-100*0.50}{50}\right)=-\frac{\pi }{4}$             

Let, current through capacitor,

As  ${?}_2-{?}_1=\frac{\pi }{2}$  

$l=\sqrt[]{{\left(I_L\right)}^2+{\left(I_C\right)}^2}$

$=\sqrt[]{4^2+2^2}=4.47A$              

Power is maximum at resonance

So, $X_L=X_C=\frac{1}{\omega C}$ ${}_{}{}_{}\frac{}{}$

$\Rightarrow C=\frac{1}{\omega *X_L}$

$=\frac{1}{250*100*{\pi }^2}$

$=4\mu F\left[{\pi }^2=10\right]$

Purely inductive circuit

$\theta =\frac{\pi }{2}$

$cos\frac{\pi }{2}=0$

Average power = 0